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I'm currently making a power/current consumption budget. Is there a difference between the quiescent current and supply current? I know the quiescent current is the current that when there is no load and is at a steady state. In comparison, the supply current is the current that is drawn from the IC no matter what.

I ask this because from this datasheet: https://www.ti.com/lit/ds/symlink/opa196.pdf?HQS=TI-null-null-digikeymode-df-pf-null-wwe&ts=1593049757175, I couldn't find the supply current for this. Do I just take the quiescent current as I normally see one or the other in a datasheet?

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  • \$\begingroup\$ Basically, I supply = Iq + Iout. For example if the output of the op-amp is connected to a 1k resistor, and it holds it at 5V, then that is 5mA which needs to be accounted for somewhere. Some current will flow in any feedback resitors you have also. \$\endgroup\$
    – mkeith
    Jun 25 '20 at 3:00
  • \$\begingroup\$ But if you tally the power in the resistors separately, then you can just use Iq for the op-amp. It is just accounting. You can have multiple categories but make sure every expense is recorded exactly once. \$\endgroup\$
    – mkeith
    Jun 25 '20 at 3:03
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If the output is sourcing current, you would need to add that to the quiescent current to get a pretty good estimate of the total current into the positive supply terminal. If the output is sinking current, then it won't (much) affect the current drawn by the positive supply terminal, but it will add to the current flowing out of the negative supply terminal.

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  • \$\begingroup\$ In my case, the IC in my original post is used as a filter, to clean data from a sensor, so that would mean the output is sinking current. Could you clarify what you mean by adding to the negative supply terminal? How I've done it so far is that the voltage supplied is +/-12V, so that is 24V total and the quiescent current is 140uA per amplifier and there is 2 amplifiers in the IC so, that is 24V*140uA. \$\endgroup\$
    – kurcio
    Jun 25 '20 at 2:29
  • \$\begingroup\$ Using the op-amp in a filter doesn't imply the output will be sinking current. That makes no sense at all. What determines the direction of output current is the output voltage, how the load is connected, and whatever current is required through the feedback path. \$\endgroup\$
    – The Photon
    Jun 25 '20 at 2:38

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